Merge branch 'master' of https://github.com/azl397985856/leetcode

Author: luzhipeng

daily/2019-06-04.md

@@ -52,7 +52,7 @@ let start = 0;
 
 for(let i = 0; i < n; i++){
     total += gas[i];
-    total += cost[i]
+    total -= cost[i]
 
     remain += gas[i];
     remain -= cost[i];

problems/283.move-zeroes.md

@@ -31,6 +31,9 @@ Minimize the total number of operations.
 无
 
 ## 代码
+* 语言支持：JS，C++
+
+JavaScript Code:
 ```js
 /*
  * @lc app=leetcode id=283 lang=javascript
@@ -79,4 +82,26 @@ var moveZeroes = function(nums) {
     }
 };
 ```
-
+C++ Code：
+> 解题思想与上面JavaScript一致，做了少许代码优化（非性能优化，因为时间复杂度都是O(n)）：
+> 增加一个游标来记录下一个待处理的元素的位置，这样只需要写一次循环即可。
+```C++
+class Solution {
+public:
+    void moveZeroes(vector<int>& nums) {
+        vector<int>::size_type nonZero = 0;
+        vector<int>::size_type next = 0;
+        while (next < nums.size()) {
+            if (nums[next] != 0) {
+                // 使用std::swap()会带来8ms的性能损失
+                // swap(nums[next], nums[nonZero]);
+                auto tmp = nums[next];
+                nums[next] = nums[nonZero];
+                nums[nonZero] = tmp;
+                ++nonZero;
+            }
+            ++next;
+        } 
+    }
+};
+```

problems/328.odd-even-linked-list.md

@@ -36,6 +36,10 @@ The first node is considered odd, the second node even and so on ...
 - 循环的结束条件设置为 `odd && odd.next && even && even.next`, 不应该是`odd && even`, 否则需要记录一下奇数节点的最后一个节点，复杂了操作
 
 ## 代码
+
+* 语言支持：JS，C++
+
+JavaScript Code:
 ```js
 /*
  * @lc app=leetcode id=328 lang=javascript
@@ -122,3 +126,30 @@ var oddEvenList = function(head) {
 };
 ```
 
+C++ Code：
+```C++
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* oddEvenList(ListNode* head) {
+        if (head == nullptr) return head;
+        auto odd = head, evenHead = head->next, even = head->next;
+        // 因为“每次循环之后依然保持odd在even之前”，循环条件可以只判断even和even->next是否为空，修改odd和even的指向的操作也可以简化
+        while (even != nullptr && even->next != nullptr) {
+            odd->next = even->next;
+            odd = odd->next;
+            even->next = odd->next;
+            even = even->next;
+        }
+        odd->next = evenHead;
+        return head;
+    }
+};
+```

problems/78.subsets.md

@@ -49,6 +49,9 @@ Output:
 
 ## 代码
 
+* 语言支持：JS，C++
+
+JavaScript Code:
 ```js
 
 /*
@@ -104,6 +107,27 @@ var subsets = function(nums) {
     return list;
 };
 ```
+C++ Code：
+```C++
+class Solution {
+public:
+    vector<vector<int>> subsets(vector<int>& nums) {
+        auto ret = vector<vector<int>>();
+        auto tmp = vector<int>();
+        backtrack(ret, tmp, nums, 0);
+        return ret;
+    }
+    
+    void backtrack(vector<vector<int>>& list, vector<int>& tempList, vector<int>& nums, int start) {
+        list.push_back(tempList);
+        for (auto i = start; i < nums.size(); ++i) {
+            tempList.push_back(nums[i]);
+            backtrack(list, tempList, nums, i + 1);
+            tempList.pop_back();
+        }
+    }
+};
+```
 
 ## 相关题目
 

problems/88.merge-sorted-array.md

@@ -156,12 +156,12 @@ var merge = function(nums1, m, nums2, n) {
     if (n === 0) return;
 
     // 为了方便大家理解，这里代码有点赘余
-    if (m < 0) {
+    if (m < 1) {
       nums1[current--] = nums2[--n];
       continue;
     }
 
-    if (n < 0) {
+    if (n < 1) {
       nums1[current--] = nums1[--m];
       continue;
     }
@@ -175,3 +175,26 @@ var merge = function(nums1, m, nums2, n) {
   }
 };
 ```
+模仿上述代码，思路相同
+C++ code:
+```
+class Solution {
+public:
+    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
+        int current = nums1.size()- 1;
+        while (current >= 0) {
+            if (n == 0) return;
+            if (m < 1) {
+                nums1[current--] = nums2[--n];
+                continue;
+            }
+            if (n < 1) {
+                nums1[current--] = nums1[--m];
+                continue;
+            }
+            if (nums1[m - 1] > nums2[n - 1]) nums1[current--] = nums1[--m];
+            else nums1[current--] = nums2[--n];
+        }
+    }
+};
+```